Crystallization to the Square Lattice for a Two-Body Potential

Abstract

We consider two-dimensional zero-temperature systems of N particles to which we associate an energy of the form

$$\begin{aligned} \mathcal {E}[V](X):=\sum _{1\leqq i<j\leqq N}V(|X(i)-X(j)|), \end{aligned}$$

where $X(j)\in \mathbb R^2$ represents the position of the particle j and $V(r)\in \mathbb R$ is the pairwise interaction energy potential of two particles placed at distance r. We show that under suitable assumptions on the single-well potential V, the ground state energy per particle converges to an explicit constant $\overline{\mathcal E}_{\mathrm {sq}}[V]$, which is the same as the energy per particle in the square lattice infinite configuration. We thus have

$$\begin{aligned} N{\overline{\mathcal E}_{\mathrm {sq}}[V]}\leqq \min _{X:\{1,\ldots ,N\}\rightarrow \mathbb R^2}\mathcal E[V](X)\leqq N{\overline{\mathcal E}_{\mathrm {sq}}[V]}+O(N^{\frac{1}{2}}). \end{aligned}$$

Moreover $\overline{\mathcal E}_{\mathrm {sq}}[V]$ is also re-expressed as the minimizer of a four point energy. In particular, this happens if the potential V is such that $V(r)=+\infty$ for$r<1$, $V(r)=-1$ for $r\in [1,\sqrt{2}]$, $V(r)=0$ if $r>\sqrt{2}$, in which case ${\overline{\mathcal E}_{\mathrm {sq}}[V]}=-4$. To the best of our knowledge, this is the first proof of crystallization to the square lattice for a two-body interaction energy.

Appendices

Appendix A. Proof of Lemmas 3.5 and 3.8

Since many constants are introduced throughout this section, for not confusing with the other constants above, we often replace $\alpha$ with $\varepsilon$. For every $\varepsilon >0$ and for every $\Xi$, $X:\Xi \rightarrow \mathbb R^2$, we recall that

$$\begin{aligned} \mathcal S_\varepsilon =\{\{p,q\}\,:\,p,q\in \Xi ,\,|x_p-x_q|\in (1-\varepsilon ,\sqrt{2}+\varepsilon )\}, \end{aligned}$$

where $x_p:=X(p)$ for every $p\in \Xi$. We first prove two preliminary Lemmas that will be useful in the proof of Lemma 3.5.

Lemma A.1

There exists $\varepsilon '>0$ such that for every $\varepsilon \in (0,\varepsilon ')$ the following holds. Let X satisfy (3.2) with $r_{\min }=1-\varepsilon$. Let $p_1,p_2,p_3\in \Xi$ and set $x_i:=X(p_i)$ for all $i\in \{1,2,3\}$. Then,

• (i) if $\{p_1,p_2\}\in \mathcal {S}_\varepsilon$ and $\{p_1,p_3\}\in \mathcal {S}_\varepsilon$ but $\{p_2,p_3\}\not \in \mathcal {S}_\varepsilon$, then in the triangle $\{x_1,x_2,x_3\}$, the interior angles satisfy $\hat{x}_1 \geqq 60^\circ$ and $\hat{x}_2,\hat{x}_3 \leqq \arccos \left( \frac{1}{2\sqrt{2}}\right) +O(\varepsilon )$;

• (ii) if all pairs amongst $p_1,p_2,p_3$ are in $\mathcal {S}_\varepsilon$, then the interior angles of the triangle $\{x_1,x_2,x_3\}$ are in the interval

  $$\begin{aligned} { \left[ \arccos \left( \frac{3}{4}\right) -O(\varepsilon ),\quad 90^\circ +O(\varepsilon ) \right] }; \end{aligned}$$

• (iii) if $\{p_1,p_2\},\{p_1,p_3\}\in \mathcal S_\varepsilon$, $|x_1-x_2|=1+O(\varepsilon )$ but $\{p_2,p_3\}\not \in \mathcal {S}_\varepsilon$, then in the triangle $\{x_1,x_2,x_3\}$, we have $\hat{x}_1\geqq \arccos \left( \frac{1}{2\sqrt{2}}\right) +O(\varepsilon )$;

• (iv) if all pairs amongst $p_1,p_2,p_3$ are in $\mathcal {S}_\varepsilon$ and $|x_1-x_2|= 1+O(\varepsilon )$, then

  $$\begin{aligned} \widehat{x_1x_2x_3},\widehat{x_3x_1x_2}\geqq {45^\circ -O(\varepsilon ).} \end{aligned}$$

  (A.1)

Proof

We set $a:=|x_1-x_2|,\, b:=|x_1-x_3|,\, c:=|x_2-x_3|$. We may assume, up to relabelling the points, that

$$\begin{aligned} 1-\varepsilon<a\leqq b<\sqrt{2} -\varepsilon . \end{aligned}$$

(A.2)

Proof of (i). The statement follows from the law of cosines, which states that

$$\begin{aligned} \cos (\hat{x}_1)=\frac{a^2+b^2-c^2}{2ab}. \end{aligned}$$

(A.3)

If $\{p_2,p_3\}\not \in \mathcal {S}_\varepsilon$, then due to (3.2) we need to have $c\geqq \sqrt{2}+\varepsilon >b$, and from (A.3) and (A.2) we find $\cos (\hat{x}_1)< a/2b\leqq 1/2$ and thus $\hat{x}_1\geqq 60^\circ$. For bounding $\hat{x}_2$, we observe that

$$\begin{aligned} { \inf }\left\{ \frac{a^2+c^2-b^2}{2ac}:\ a,b\in (1-\varepsilon ,\sqrt{2}+\varepsilon ),\ c\geqq \sqrt{2} + \varepsilon \right\} \end{aligned}$$

is reached as $(a,b,c)\rightarrow (1-\varepsilon ,\sqrt{2}+\varepsilon ,\sqrt{2}+\varepsilon )$, and equals the value of the expression $(a^2+c^2-b^2)/(2ac)$ in that limit, giving the desired bound on $\hat{x}_2$. The bound for $\hat{x}_3$ works similarly, with the roles of a, b interchanged.

Proof of (ii). If $\{p_2,p_3\}\in \mathcal {S}_\varepsilon$ then $c\in (1-\varepsilon ,\sqrt{2}+\varepsilon )$. Moreover (A.2) holds. In such a range, the sup of the right hand side of (A.3) is realized by $c=1-\varepsilon ,\, a=b=\sqrt{2} + \varepsilon$, in which case

$$\begin{aligned} \hat{x}_1=\arccos \left( \frac{3+(4\sqrt{2}+2)\varepsilon +\varepsilon ^2}{2(\sqrt{2}+\varepsilon )^2} \right) =\arccos \left( \frac{3}{4}+O(\varepsilon )\right) =\arccos \left( \frac{3}{4}\right) +O(\varepsilon ), \end{aligned}$$

whereas the inf is reached for $a=b=1-\varepsilon ,\, c=\sqrt{2}+\varepsilon$, in which case

$$\begin{aligned} \hat{x}_1=\arccos \left( \frac{-(4+2\sqrt{2})\varepsilon +\varepsilon ^2}{2(1-\varepsilon )^2}\right) =\arccos (-O(\varepsilon ))=90^\circ +O(\varepsilon ). \end{aligned}$$

Proof of (iii). By (A.3) and by the hypothesis we have

$$\begin{aligned} \cos (\hat{x}_1)=\frac{(1+O(\varepsilon ))^2+b^2-c^2}{2(1+O(\varepsilon ))b}\leqq \frac{b^2-1-O(\varepsilon )}{2(1+O(\varepsilon ))b}. \end{aligned}$$

It is easy to see that the quantity on the right-hand-side is - for $\varepsilon$ small enough - monotonically increasing with respect to b, so that it is maximized for $b=\sqrt{2}+\varepsilon$. From this, the claim follows.

Proof of (iv). Again by (A.3) and by the hypothesis, we have

$$\begin{aligned} \cos (\hat{x}_1)\leqq \frac{{(1+O(\varepsilon ))^2}+c^2-b^2}{2(1-\varepsilon )c}, \end{aligned}$$

where the sup of the right-hand-side is reached for $|x_2-x_3|=\sqrt{2}+\varepsilon$ and $|x_1-x_3|=1-\varepsilon$, thus yielding the claim.

By applying verbatim the same reasoning of the proof of Lemma A.1(ii) one gets the following result:

Corollary A.2

Let $X\subset \mathbb R^2$ and let $x,y,z\in X$ be such that $\{x,y\}, \{y,z\}\in \mathcal {S}_0(X)$. Then $\widehat{xyz}\geqq \arccos \left( \frac{3}{4}\right) \sim 41.4^\circ$.

Lemma A.3

There exists $\varepsilon ''\in (0,\varepsilon ']$ (with $\varepsilon '$ given by Lemma A.1) such that for every $\varepsilon \in (0,\varepsilon '')$ the following holds: let $\Xi$ be a set of labels and let $p_1,p_2,p_3,p_4\in \Xi$ be such that $\{p_i,p_j\}\in \mathcal {S}_{\varepsilon }$ for all $i\ne j$ and set $x_i:=X(p_i)$ for all $i=1,\ldots ,4$; then, up to relabeling, for all $i\in \{1,2,3,4\}$ we have

• (i) $| x_i-x_{i+1}|= 1+ O(\varepsilon )$,

• (ii) $\widehat{x_i x_{i+1}x_{i+2}}=90^\circ +O(\varepsilon )$,

• (iii) $|x_i-x_{i+2}|\leqq \sqrt{2}+O(\varepsilon )$,

where $x_{i+4}=x_{i}$ for every $i=1,\ldots ,4$. If $p_1,p_2,p_3,p_4$ are such that $\{p_i,p_j\}\in \mathcal {S}_{\varepsilon }$ for all $i\ne j$, then the quadrilateral $\{p_1,p_2,p_3,p_4\}$ is called an $\varepsilon$-square.

Proof

Up to relabeling we may suppose that the points $x_1,\ldots ,x_4$ are in cyclic order along the boundary of the convex hull $\mathrm {Conv}(\{x_1,\ldots ,x_4\})$.

Proof of (i). Assume that $|x_1-x_2|\geqq |x_i-x_{i+1}|$ for every $i=2,\ldots ,4$. Then, under the constraints $|x_i-x_{i+1}|> 1-\varepsilon$ and $|x_i-x_{i+2}|< \sqrt{2} +\varepsilon$, the sup of $|x_1-x_2|$ is realized by $|x_2-x_3|=|x_3-x_4|=|x_4-x_1|=1-\varepsilon$ and $|x_1-x_3|=|x_2-x_4|=\sqrt{2}+\varepsilon$, which gives the desired bound.

Proof of (ii). It follows directly by Lemma A.1.

Proof of (iii). By the law of cosines, (i) and (ii), we have

$$\begin{aligned} |x_i-x_{i+2}|^2{= 2(1+O(\varepsilon ))^2-2(1+O(\varepsilon ))^2=2+O(\varepsilon )}, \end{aligned}$$

which gives the claim.

We are now in a position to prove Lemma 3.5.

Proof of Lemma 3.5

We assume that $p\in \Xi$ has the maximum number of 8 neighbors $\mathcal {G}_\varepsilon$. We write $x=X(p)$ and we set $x_i=X(p_i)$ for every $i=1,\ldots ,8$. Without loss of generality the $x_i$ are ordered in counterclockwise order around x. We recall that $\arccos \left( \frac{1}{2\sqrt{2}}\right) \sim 69.2^\circ$ and $\arccos \left( \frac{3}{4}\right) \sim 41.4^\circ$. Let $\varepsilon ''$ be the constant given in Lemma A.3.

Claim 1: There exists $\varepsilon _0\in (0,\varepsilon '')$ such that for all $\varepsilon \in (0,\varepsilon _0)$ at least 7 indices $i=1,\ldots ,8$ are such that $\{p_i,p_{i+1}\}\in \mathcal {S}_{\varepsilon }$. We first note that if for more than two indices i there holds $|x_i-x_{i+1}|\geqq \sqrt{2}+\varepsilon$ then by Lemma A.1, $\widehat{x_ixx_{i+1}}\geqq 60^\circ$, and thus at least one of the remaining 6 angles is smaller than $(360^\circ -120^\circ )/6=40^\circ <\arccos \left( \frac{3}{4}\right)$ . As a consequence, there exists $\varepsilon _0>0$ such that for $\varepsilon \in [0,\varepsilon _0)$ we get a contradiction with Lemma A.1 and hence $|x_i-x_{i+1}|\geqq \sqrt{2}+\varepsilon$ may hold for at most one index $i\in \mathbb Z/8\mathbb Z$ (Fig. 5).

Fig. 5

Illustration of Claim 1. The dashed line represents the possible bond not in $\mathcal S_\varepsilon$

Claim 2: For all $\varepsilon \in (0,\varepsilon _0)$, the configuration ${X(\mathcal N_\varepsilon (p))}=\{x,x_1,\ldots ,x_8\}$ contains at least one $\varepsilon$-square as defined in Lemma A.3. Assume that this is not the case , namely that there exists $\varepsilon \in (0,\varepsilon _0)$ such that $X(\mathcal N_\varepsilon (p))$ does not contain any $\varepsilon$-square. Then, by Claim 1, each of the $p_i$’s ($i=1,\ldots ,8$) has two or three neighbors in $\mathcal N_\varepsilon (p)$. Note also that the sum of internal angles of the octagon $\{x_1,\ldots ,x_8\}$ is $1080^\circ$, thus at least one angle is larger than $1080^\circ /8=135^\circ$, say it is the angle at $\widehat{x_1x_2x_3}$. Since $\mathcal {N}_\varepsilon (p)\cap \partial \mathcal {G}_\varepsilon =\emptyset$, $x_2$ also has 8 neighbors. By considering the successive angles around $x_2$ formed with the 8 neighbors of $p_2$ in $\mathcal {G}_\varepsilon$, we have that 6 such angles are contained outside the sector spanned by the angle $\widehat{x_1x_2x_3}$, therefore at least one of these angles is smaller than or equal to $(360^\circ -135^\circ )/6=37.5^\circ <\arccos \left( \frac{3}{4}\right) {+O(\varepsilon )}$ for $\varepsilon \in [0,\varepsilon _0)$ where $\varepsilon _0$ is the one given Claim 1. But this fact contradicts Lemma A.1, and hence we get the claim (Fig. 6).

Claim 3: There exists $\varepsilon _1\in (0,\varepsilon _0]$ such that for all $\varepsilon \in (0,\varepsilon _1)$, the configuration $X(\mathcal N_\varepsilon (p))$ contains at least two $\varepsilon$-squares. Assume that this is not the case , namely that there exists a sequence $\{\varepsilon _n\}_{n\in \mathbb N}$ with $\varepsilon _n\rightarrow 0^+$ as $n\rightarrow +\infty$ such that every $n\in \mathbb N$ there exists $q_n\in \Xi$ such that $\mathcal {N}_{\varepsilon _n}(q_n)\cap \partial \mathcal {G}_{\varepsilon _n}=\emptyset$ and $X(\mathcal N_{\varepsilon _n}(X(q_n)))$ does not contain two $\varepsilon _n$-squares. Fix $n\in \mathbb N$, and let $\varepsilon =\varepsilon _n$ and $p=q_n$ be as above. In view of Claim 2, this means that $X(\mathcal {N}_{\varepsilon }(p))$ contains only one $\varepsilon$-square. Let $\{x,x_1,x_2,x_3\}$ be such $\varepsilon$-square.

Fig. 6

Hypothesis and final situation reached by Claim 2

If $|x_1-x_8|\geqq \sqrt{2}+\varepsilon$ then by Lemma A.1(iii), $\widehat{x_8xx_1} \geqq 69^\circ$ for $\varepsilon$ sufficiently small. But in this case, by Lemma A.3(ii) we conclude that

$$\begin{aligned} \sum _{i=4}^{8}\widehat{x_{i-1}xx_{i}}\leqq 360^\circ -69^\circ -90^\circ {+O(\varepsilon )}=201^\circ +{O(\varepsilon )}, \end{aligned}$$

which implies that the smallest angle between the $\widehat{x_{i-1}xx_i}$, for $4\leqq i\leqq 8$, is smaller than $40.2^\circ <\arccos \left( \frac{3}{4}\right)$, thus contradicting Lemma A.1 for $\varepsilon$ small enough.

This shows that for $\varepsilon _1>0$ sufficiently small we have $\{p_8,p_1\}\in \mathcal S_\varepsilon$. Similarly we find $\{p_3,p_4\}\in \mathcal S_\varepsilon$.

By Lemma A.1, we have $\widehat{xx_3x_4}, \widehat{xx_1x_8}\leqq 90^\circ +{O(\varepsilon )}$, and as before, at least one of the 5 remaining internal angles of the octagon $\{x_1,\ldots ,x_8\}$ at vertices $x_4,x_5,x_6,x_7,x_8$ is larger than or equal to

$$\begin{aligned} \theta _\varepsilon :=\frac{1}{5}\left( 1080^\circ -450^\circ -{O(\varepsilon )}\right) =126^\circ -{O(\varepsilon )}. \end{aligned}$$

Say that $x_i$ is such a vertex. We are under the assumption that no $\varepsilon$-square at x contains $x_i$, thus by considering the possible allowed $\mathcal S_\varepsilon$-edges between vertices in $\mathcal N_\varepsilon (p)$ we find that $\sharp \left( \mathcal N_\varepsilon (p_i)\cap \mathcal N_\varepsilon (p)\setminus \{p_i\}\right) \leqq 3$. On the other hand, we are also under the assumption that $\mathcal {N}_\varepsilon (p)\cap \partial \mathcal {G}_\varepsilon =\emptyset$, thus $\sharp (\mathcal N_\varepsilon (p_i)\setminus \{p_i\})=8$. Thus there are 6 angles at $x_i$ formed by successive neighbors of $x_i$ and not contained in $\widehat{x_{i-1}x_ix_{i+1}}$. At least one of these angles is smaller than or equal to

$$\begin{aligned} \beta _\varepsilon :=(360^\circ -\theta _\varepsilon )/6\leqq {39^\circ }+{O(\varepsilon )}. \end{aligned}$$

For $\varepsilon _0$ small enough we find $\beta _\varepsilon <\arccos (\frac{3}{4}) +O(\varepsilon )$, contrary to Lemma A.1, and our claim follows.

Claim 4: There exists $\varepsilon _2\in (0,\varepsilon _1]$ such that for all $\varepsilon \in [0,\varepsilon _2)$ the configuration $X\cap B(x,\sqrt{2}+\varepsilon )$ cannot contain only two $\varepsilon$-squares with no common edges and two further successive edges from x. We will call “remaining vertices” the nearest neighbors of x that do not belong to an $\varepsilon$-square. From Claim 3, we know that there is at most 2 remaining vertices for $\varepsilon <\varepsilon _1$. Again we prove the claim by contradiction. Up to cyclic relabeling of the $x_i$ the two $\varepsilon$-squares are $\{x,x_1,x_2,x_3\}$ and $\{x,x_4,x_5,x_6\}$. By Lemma A.3 and by the law of cosines we obtain

$$\begin{aligned} \widehat{x_3 x x_{4}}\geqq \arccos \left( \frac{2(1+O(\varepsilon ))^2-(1-\varepsilon )^2}{2(1+O(\varepsilon ))^2}\right) =\arccos \left( \frac{1}{2}+O(\varepsilon )\right) =60^\circ +O(\varepsilon ). \end{aligned}$$

Moreover, by using again Lemma A.3, at least one of the angles $\widehat{x_ixx_{i+1}}, i=6,7,8$ must be smaller than or equal to

$$\begin{aligned} \xi _\varepsilon :=\frac{1}{3}(360^\circ -2({90^\circ +O(\varepsilon )})-60^\circ -O(\varepsilon ))=40^\circ +O(\varepsilon )<\arccos \left( \frac{3}{4}\right) +O(\varepsilon ), \end{aligned}$$

for $\varepsilon$ small enough. Therefore a contradiction to Lemma A.1 follows (Fig. 7).

Fig. 7

What Claim 4 proves impossible

Claim 5: There exists $\varepsilon _3\in (0,\varepsilon _2]$ such that for all $\varepsilon \in (0,\varepsilon _3)$ the following holds: if the configuration $X\cap B(x,\sqrt{2}+\varepsilon )$ contains two $\varepsilon$-squares with no common edges and the two remaining vertices that are not successive, then it contains a further $\varepsilon$-square, sharing one edge with each given $\varepsilon$-squares.

Fig. 8

Hypothesis and end result of Claim 5

Up to cyclic relabeling of the $x_i$’s, the two $\varepsilon$-squares are $\{x,x_1,x_2,x_3\}$ and $\{x,x_5,x_6,x_7\}$. We can assume without loss of generality $\beta _4:=\widehat{x_3 x x_5}\leqq \widehat{x_7 x x_1}=:\beta _8$, $\beta _4^-:=\widehat{x_3 x x_4}\leqq \widehat{x_4 x x_5}=:\beta _4^+$, and $\beta _8^-:=\widehat{x_7 x x_8}\leqq \widehat{x_8 x x_1}=:\beta _8^+$. By Lemma A.3, it follows that $\beta _4\leqq 90^\circ -O(\varepsilon )$, $\beta _8\geqq 90^\circ -O(\varepsilon )$, and $\beta _4^-\leqq 45^\circ -O(\varepsilon )$. By the assumption $|x_4-x|\leqq \sqrt{2} +\varepsilon$ . By the law of cosines, we have

$$\begin{aligned} \begin{aligned} (1-\varepsilon )^2\leqq&|x_3-x_4|^2=|x_3-x|^2+|x_4-x|^2-2 |x_3-x||x_4-x|\cos \beta _4^-\\ \leqq&(1+O(\varepsilon ))^2+|x_4-x|^2-2(1+O(\varepsilon ))|x_4-x|\cos \left( 45^\circ +{O(\varepsilon )}\right) , \end{aligned} \end{aligned}$$

(A.4)

whence, using

$$\begin{aligned} \cos \left( 45^\circ +O(\varepsilon )\right) =\frac{\sqrt{2}}{2}+O(\varepsilon )\quad \text { and }\quad (1+O(\varepsilon ))^2-(1-\varepsilon )^2=O(\varepsilon ), \end{aligned}$$

we deduce the following inequality

$$\begin{aligned} |x_4-x|^2-({\sqrt{2}}+O(\varepsilon )) |x_4-x|+O(\varepsilon )\geqq 0; \end{aligned}$$

it follows that $|x_4-x|=\sqrt{2}+O(\varepsilon )$.

Moreover, by (A.4), it follows also that, for $\varepsilon$ small enough,

$$\begin{aligned} \begin{aligned} \cos \beta _4^-=&\frac{|x_3-x|^2+|x_4-x|^2-|x_4-x_3|^2}{2 |x_3-x||x_4-x|}\\ \leqq&\frac{(1+O(\varepsilon ))^2+(\sqrt{2}+O(\varepsilon ))^2-(1-\varepsilon )^2}{2(1-\varepsilon )(\sqrt{2}+O(\varepsilon ))}=\frac{\sqrt{2}}{2}+O(\varepsilon ), \end{aligned} \end{aligned}$$

which, together with the assumption on $\beta _4^-$, implies that $\beta _4^-=45^\circ +O(\varepsilon )$. Using again the law of cosines one can easily deduce that $|x_3-x_4|=1+O(\varepsilon )$ and that $\widehat{x x_3 x_4}=90^\circ +O(\varepsilon )$. Analogously, one can see that $\beta _4^+=45^\circ +O(\varepsilon )$ and that $|x_4-x_5|=1+O(\varepsilon )$. It follows that $|x_3-x_5|=\sqrt{2}+O(\varepsilon )$. Finally, since $p_3$ has 8 neighbors, arguing by contradiction one can show that $|x_3-x_5|\leqq \sqrt{2}+\varepsilon$. In conclusion, $\{x,x_3,x_4,x_5\}$ is an $\varepsilon$-square and then the Claim follows (Fig. 8).

Claim 6: There exists $\varepsilon _4\in (0,\varepsilon _3]$ such that for all $\varepsilon \in (0,\varepsilon _4)$ the following holds: $X(\mathcal N(p))$ contains at least 3 adjacent $\varepsilon$-squares. In view of Claims 3 and 5, the Claim needs to be proven only in the case that there are two $\varepsilon$-squares sharing one edge. Let $\{x,x_1,x_2,x_3\}$ and $\{x,x_3,x_4,x_5\}$ be two $\varepsilon$-squares. By Lemma (A.3), we have that

$$\begin{aligned} \sum _{j=1}^3\widehat{x_{j}x_{j+1}x_{j+2}}\leqq 360^\circ + O(\varepsilon )\,,\qquad \widehat{xx_1x_2}+\widehat{x_4x_5x}\leqq 180^\circ +O(\varepsilon ), \end{aligned}$$

(A.5)

whereas, by Lemma A.1, we obtain

$$\begin{aligned} \widehat{x_8x_1x}+\widehat{xx_5x_6}\leqq 180^\circ +O(\varepsilon ). \end{aligned}$$

(A.6)

Then, using again that the sum of the internal angles of the octagon is $1080^\circ$, we have

$$\begin{aligned} \widehat{x_5x_6x_7}+\widehat{x_6x_7x_0}+\widehat{x_7x_0x_1}\geqq 1080^\circ -720^\circ -O(\varepsilon )=360^\circ - O(\varepsilon )\,. \end{aligned}$$

Therefore, one of the above three angles, say $\widehat{x_{i-1}x_{i}x_{i+1}}$ is larger than

$$\begin{aligned} \vartheta _\varepsilon =120^\circ -O(\varepsilon ). \end{aligned}$$

Since $p_i$ has eight neighbors in $\mathcal {G}_\varepsilon$ and since $x_i$ does not belong to an $\varepsilon$-square, $p_i$ has exactly three neighbors in $\mathcal N_\varepsilon (p)$ and their images through X cover an angle at $x_i$ of at least $\vartheta _{\varepsilon }$ . Therefore amongst the remaining 6 angles at $x_i$ spanned by successive neighbors of $x_i$, at least one is smaller than or equal to

$$\begin{aligned} \frac{360^\circ -\vartheta _{\varepsilon }}{6}=40^\circ +O(\varepsilon ), \end{aligned}$$

contradicting Lemma A.1, and concluding the proof of Claim 6.

Claim 7: There exists $\varepsilon _5\in (0,\varepsilon _4]$ such that for all $\varepsilon \in (0,\varepsilon _5)$ $X(\mathcal {N}_\varepsilon (p))$ contains four $\varepsilon$-squares.

By Claim 6, we can assume that there are three $\varepsilon$-squares. Up to relabeling such $\varepsilon$-squares are $\{x,x_1,x_2,x_3\}$, $\{x,x_3,x_4,x_5\}$, and $\{x_5,x_6,x_7,x \}$. By Lemma A.3 we have

$$\begin{aligned} \widehat{x_7xx_1}=90^\circ +O(\varepsilon ), \end{aligned}$$

(A.7)

and hence, by the law of cosines,

$$\begin{aligned} |x_7-x_1|=\sqrt{2}+O(\varepsilon ). \end{aligned}$$

Moreover, again by the law of cosines the remaining angles $\widehat{x_7 x x_0}, \widehat{x_0 x x_1}$ also are $O(\varepsilon )$-close to $45^\circ$. By arguing as in Claim 5 one can easily get the claim.

Set $\alpha _0:=\varepsilon _5$. In view of Claim 7 and of the very definition of $\varepsilon$-square, (3.17) is satisfied for $\varepsilon \in [0,\alpha _0)$. We therefore define $\phi :\mathcal {N}_\varepsilon (p)\rightarrow \{-1,0,1\}^2$ as in (3.16) and by all the Claims above, it is easy to show that $\delta _\phi (x',x'')\leqq C_3 \alpha |x'-x''|$ for all $x',x''\in \{x,x_1,\ldots ,x_8\}$ for some constant $C_3\in [1,\frac{1}{\alpha _0})$ (depending only on $\alpha _0$).

Notice that for $\varepsilon =0$ the $\varepsilon$-squares are nothing but the unit squares. Therefore, by the same proof as for Lemma 3.5 with $\alpha =0$ we obtain the following result:

Corollary A.4

Let $X\in \mathcal {C}$ and let $x\in X$ have 8 neighbors in $\mathcal {G}_0(X)$, each of which has in turn 8 neighbors in $\mathcal {G}_0(X)$. Let $x_1,\ldots ,x_8,x_9\equiv x_1$ be the neighbors of x ordered counterclockwise around x and let $|x_1-x|=\min _{i=1,\ldots ,8}|x-x_i|$. Then, the quadrilaterals $\{x,x_1,x_{2},x_{3}\}$, $\{x,x_3,x_{4},x_{5}\}$, $\{x,x_5,x_{6},x_{7}\}$, $\{x,x_7,x_{8},x_{9}\}$ are all unit squares.

We next pass to proving Lemma 3.8.

Proof of Lemma 3.8

Set $\alpha '_0:=\varepsilon ''$ where $\varepsilon ''$ is the one given in Lemma A.3. Let $\alpha \in (0,\alpha _0']$ and let $\{p_1,p_2,p_3,p_4\}$ denote the set of vertices of $\mathcal G_\alpha$. By hypothesis, $\{p_i,p_j\}\in \mathcal {S}_\alpha$ for every $i,j=1,\ldots ,4$ with $i\ne j$. Then, the assumptions of Lemma A.3 are satisfied (with $\varepsilon$ replaced by $\alpha$), so that setting $x_i:=X(p_i)$ for every $i=1,\ldots ,4$ and $x_{i+4}\equiv x_i$ for every $i\in \mathbb Z$, we deduce that, up to a relabeling, (i),(ii), and (iii) hold true. In particular, for every $\alpha \in [0,\alpha '_0)$ we have

$$\begin{aligned} 1-\varepsilon \leqq |x_{i+1}-x_{i}|=1+O(\alpha ),\qquad \sqrt{2}+O(\alpha )=|x_{i}-x_{i+2}|\leqq \sqrt{2}+\alpha . \end{aligned}$$

(A.8)

Therefore, by (A.8), there exists a constant $C_3'\in [1,\frac{1}{\alpha '_0})$ (depending only on $\alpha '_0$) and a map $\phi :\{p_1,p_2,p_3,p_4\}\rightarrow \{0,1\}^2$ with

$$\begin{aligned} \phi (p_1)=(0,0), \quad \phi (p_2)=(1,0), \quad \phi (p_3)=(1,1), \quad \phi (p_4)=(0,1), \end{aligned}$$

such that $\delta _{\phi }(x',x'')\leqq C_3\alpha |x'-x''|$ for all $x',x''\in \{x_1,\ldots ,x_4\}$. As a consequence, (3.19) holds true.

Appendix B. List of Notations

Below we produce a list of those notations used at several points in the paper which we feel would help the reader orient, together with the main equations in the paper in which those notations are introduced.

$\mathcal E[V](X_N)$	Energy defined in (1.1)
$\overline{\mathcal E}_{\mathrm {sq}}[V]$	Minimal energy per point of a square lattice, see (1.4)
$\mathcal E_4[V](x_1,x_2,x_3,x_4)$	Energy as in (1.6) (see also Fig. 1) and later also (3.21)
$E_\beta , E_\beta ^1, E_\beta ^2$	Intervals of distances, see (1.7)
W(s)	(1.11), and later also (3.20)
$\mathcal Z_\boxtimes$	Combinatorial model-space from (3.10)
$\mathcal S_{\alpha }, \mathcal G_{\alpha }, \mathcal N_\alpha (p), \partial \mathcal G_{\alpha }$	Graph data from (3.11)
$X_1\sim _\alpha X_2$	$\alpha$-deformation, see (3.15)
$\mathcal X_4(\mathbb R^2)/\mathrm {Isom}(\mathbb R^2)$	Space of 4-point configurations first appearing in (3.26)
$\mathscr {Q}$	Configuration forming the vertices of a convex quadrilateral ordered along its perimeter, see Subsection 3.4.1
$\mathscr {Q}_\alpha$	Small deformations of squares, see (3.28)
$\mathscr {S}_\alpha$	Small dilations of squares, see (3.29)
$E_\alpha ^{sq}$	Square distances corresponding to $E_\alpha$, and defined in (3.30)
$\mathcal {D}$	Set of distances from $\mathbb Z^2$, defined in (4.3)
$\mathrm {Ell}_\alpha (a,b)$	Ellipse defined in (4.2)
$\mathrm {Sides}(Q'_r), \mathrm {Sides}(\mathcal {Q}'_r), \mathrm {Diag}(Q'_r) \mathrm {Diag}(\mathcal {Q}'_r)$	See (4.4)
$\mathrm {Sides}(Q_r), \mathrm {Sides}(\mathcal {Q}_r), \mathrm {Diag}(Q_r), \mathrm {Diag}(\mathcal {Q}_r)$	See (4.5)
$\mathcal {L}_r$	Sublattices of $\mathbb Z^2$ of scale r, defined in (4.16a)
$\quad m(r)$	Multiplicities of sublattices, defined in (4.16b)
$\widetilde{\mathcal D}\subset \mathcal D$	The subset constructed in Lemma 4.12
$\mathcal {NQ}^{(1)}, \mathcal {NQ}^{(2)}, \mathcal {NQ},$	Sets defined in (4.27), (4.28), and (4.43) respectively
$\mathcal {Q}^{b}_{r}(Q_1)$	Squares of scale r intersecting $Q_1$, see (4.29)
e(v, r)	Error term as in (4.36)
$\widetilde{W}(t^2), W_*(t^2)$	Resummed interaction potentials defined in (4.38)
$\mathrm {err}_1(r,V), \mathrm {err}_2(r,V)$	Error terms from (4.47)
$\mathrm {err}_3(r,V)$	Error term from (4.52)
$\mathrm {err}_4(V,X)$	Error term from (4.59)
$V_{**}(r)$	Interaction potential defined in (4.63)